Question

Scores for a common standardized college aptitude test are
normally distributed with a mean of 493 and a standard deviation of
115. Randomly selected men are given a Test Preparation Course
before taking this test. Assume, for sake of argument, that the
preparation course has no effect.

If 1 of the men is randomly selected, find the probability that his
score is at least 545.2.

P(*X* > 545.2) =

Answer as a number accurate to 4 decimal places.

If 14 of the men are randomly selected, find the probability that
their mean score is at least 545.2.

P(x > 545.2) =

Answer as a number accurate to 4 decimal places.

Assume that any probability less than 5% is sufficient evidence to
conclude that the preparation course does help men do better. If
the random sample of 14 men does result in a mean score of 545.2,
is there strong evidence to support the claim that the course is
actually effective?

Answer #1

**1)**

P(X > 545.2) = P( (X - mean ) / SD > (545.2 - mean) / SD )

= P(Z > (545.2 - 493) / 115)

= P( Z > 0.454)

= 1 - P( Z < 0.454)

= 1 - 0.6751

= **0.3249**

P(X > 545.2) = **0.3249**

**2)**

P(X > 545.2) = P( (X - mean ) / (SD / n^{0.5}) >
(545.2 - mean) / (SD / n^{0.5}) )

= P(Z > (545.2 - 493) / (115 / 14^{0.5}))

= P( Z > 1.698)

= 1 - P( Z < 1.698)

= 1 - 0.9552

= **0.0447**

P(X > 545.2) = **0.0447**

**3)**

Since Probability = 0.0447 < 0.05 ( 5% ) , we can conclude
that **the preparation course does help men do better and
there is strong evidence to support the claim that the course is
actually effective.**

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